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3 years ago

A 500-kg elevator is raised to a height of 20 m in 10 seconds. Calculate the power of the motor.

Physics

2 answers:

LuckyWell [14K]3 years ago

4 0

<>"power = 500 kg x 9.8 m/s² x 20 m / 10s = 9800 watts."<>

lyudmila [28]3 years ago

3 0

Answer:

Power, P =9800 W

Explanation:

It is given that,

Mass of the elevator, m = 500kg

It is raised to a height of 20 m in 10 seconds

Power is defined as the work done per unit time.

$P=\dfrac{W}{t}$

Work done is given by :

$W=mgh$

h is the height

So, $P=\dfrac{mgh}{t}$

$P=\dfrac{500\ kg\times 9.8\ m/s^2\times 20\ m}{10\ s}$

P = 9800 Watts

Hence, the power of the motor is 9800 W. The correct option is (b).

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A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42◦ from t

Semmy [17]

Answer:

A) P.E = 138.44 J

B) The velocity of swing at bottom, v = 3.33 m/s

C) The work done, W = -138.44 J

Explanation:

Given,

The mass of the child, m = 25 Kg

The length of the swing rope, L = 2.2 m

The angle of the swing to the vertical position, ∅ = 42°

A) The potential energy at the initial position ∅ = 42° is given by the relation

P.E = mgh joule

Considering h = 0 for the vertical position

The h at ∅ = 42° is h = L (1 - cos∅)

P.E = mgL (1 - cos∅)

Substituting the given values in the above equation

P.E = 25 x 9.8 x 2.2 (1 - cos42°)

= 138.44 J

The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J

B) The velocity of the swing at the bottom.

At bottom of the swing the P.E is completely transformed into the K.E

∴ K.E = P.E

1/2 mv² = 138.44

1/2 x 25 x v² 138.44

v² = 11.0752

v = 3.33 m/s

The velocity of the swing at the bottom is, v = 3.33 m/s

C) The work done by the tension in the rope from initial position to the bottom

Tension on string, T = Force acting on the swing, F

$W=L\int\limits^0_\phi{F} \, d \phi$

$=L\int\limits^0_\phi{mg.sin \phi} \, d \phi$

= $-Lmg[cos\phi]_{42}^{0}$

= - 2.2 x 25 x 9.8 [cos0 - cos 42°]

= - 138.44 J

The negative sign in the in energy is that the work done is towards the gravitational force of attraction.

The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J

3 0

3 years ago

Why are you able to observe the Doppler effect on earth with sound waves but not with light waves?

VladimirAG [237]

Answer: When an ambulance passes with its siren blaring, you hear the pitch of the siren change: as it approaches, the siren’s pitch sounds higher than when it is moving away from you. This change is a common physical demonstration of the Doppler effect.

Explanation:

6 0

3 years ago

Two identical satellites are in orbit about the earth. One orbit has a radius r and the other 2r. The centripetal force on the s

velikii [3]

Answer:

the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.

Explanation:

Mass of satellite, m

orbit radius of first, r1 = r

orbit radius of second, r2 = 2r

Centripetal force is given by

$F= \frac{mv^{2}}{r}$

Where v be the orbital velocity, which is given by

$v=\sqrt{gr}$

So, the centripetal force is given by

$F= \frac{mgr}}{r}}=mg$

where, g bet the acceleration due to gravity

$g=\frac{GM}{r^{2}}$

So, the centripetal force

$F= \frac{GMm}}{r^{2}}}$

Gravitational force on the satellite having larger orbit

$F= \frac{GMm}{4r^{2}}$ .... (1)

Gravitational force on the satellite having smaller orbit

$F'= \frac{GMm}{r^{2}}$ .... (2)

Comparing (1) and (2),

F' = 4 F

So, the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.

8 0

3 years ago

What is Kinetic Energy based on?

jolli1 [7]

Answer:

C. Motion

Explanation:

Hope this helps!

3 0

3 years ago

1) A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.5 m/

mojhsa [17]

Answer: a) 49.560 and 21.13 b) i) 50 N, ii) 196 N iii) 196 N iv) 47.685 N

c) i) 594.72 ii) 0 iii) 0 iv) 0

d) 594.72

Explanation: question a)

The force is inclined at an angle of 25° to the horizontal

The horizontal component of force = 50 cos 25° = 49.560 N

The vertical component of force = 50 sin 30°= 21.130N

Question b)

i) according to the question applied force is 50 N

ii) if g = 9.8m/s², w=mg where m = mass of object = 20kg hence weight = 20* 9.8 = 196 N

iii) the normal force is the force the floor exerts on the body as a result of the weight of the object.

Normal reaction R = W = mg, we already deduced that w = mg, hence R = 196 N.

iv) according to newton's laws of motion

F - Fr = ma

F = applied force = horizontal component of force = 49.560 N.

We need to get the acceleration (a) by using Newton laws of motion before we can be able to compute the frictional force..

The body started from rest hence initial velocity u = 0

Final velocity v = 1.5m/s distance covered (s) = 12m

v ² = u² + 2as

But u = 0

v² = 2as

1.5² = 2(a) * 12

2.25 = 24a

a = 2.25/24 = 0.09735m/s²

From F - Fr = ma

49.560 - Fr = 20 * 0.09735

49.560 - Fr = 1.875

Fr = 49.560 - 1.875

Fr = 47.685 N

Question c)

i) The applied force = 49.560 N, distance covered = 12m

Work done = force * distance

Work done = 49.560 * 12

Work done = 594.72 J

ii) the weight of the object does not make the object move a distance, hence work done = 0 ( since distance covered is 0)

iii) the normal force is the same thing as the weight and they did not cover any distance hence work done is zero.

iv) the frictional force does not cover any distance, hence work done is zero.

Question d)

The total work done = work done by applied force + work done by weight + work done by normal reaction + work done by frictional force.

Total work done = 594.72 + 0 + 0 + 0 = 594.72 J

8 0

2 years ago